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If CaCl_2 is dissolved in 250 ml of solution and gives 3.011 x 10^23 chloride ions, what is the molarity of the CaCl_2 solution? a) 0.12 M b) 0.24 M c) 0.48 M d) 0.96 M

User Omar Kohl
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Answer:

The CaCl2 solution is 1.0M. The closest option is d) 0.96 M. Perhaps the value of Avogadro's number has changed slightly from the value of 6.022 x 10^23 used here.

Step-by-step explanation:

It is useful to remember the definition of Molar (M). It is concentration expressed as moles/liter. 1 mole of NaCl dissolved in 1 liter of water provides a 1 M solution. 1 mole of NaCl in 500ml (0.5 liters) would make a 2 M solution (1 mole/0.5L) = 2M

In this problem, we are told 3.011 x 10^23 chloride ions are present when CaCl2 is dissolved. Avogadro's number says 6.02 x 10^23 ions (or molecules) are equal to 1 mole. We have:

(3.011 x 10^23 chloride ions)/(6.02 x 10^23 chloride ions/mole) or

0.50 mole Cl ions

Since CaCl2 releases 2 Cl ions per molecule of CaCl2, we must have started with 0.25 [0.50/2] moles of CaCl2.

We were told that the CaCl2 was dissolved in 250 ml of solution. Convert to liters: 0.250 liters.

0.25 moles in 0.250 liters means that the molarity of the solution is (0.25 moles CaCl2)/(0.25 L) = 1.0 mole/liter or 1.0 mole

User Uri Granta
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